Problem: $f(x)=\dfrac{1}{2-x}$ Find a power series for $f$. Choose 1 answer: Choose 1 answer: (Choice A) A $ 1+\frac{1}{2}x+\frac{1}{4}x^2+\ldots +\left(\frac{1}{2}\right)^n x ^n+\ldots$ (Choice B) B $ 1-x+x^2+\ldots +\left(-1\right)^n x ^n+\ldots$ (Choice C) C $ 1-x-x^2+\ldots -x ^n+\ldots$ (Choice D) D $ \frac{1}{2}+\frac{x}{4}+\frac{x^2}{8}+\ldots +\frac{x^n}{2^{n+1}}+\ldots$
Explanation: We divide the numerator and the denominator by $2\,$ so that the denominator has a constant term of $1\,$. $ f(x)= \frac{1}{2-x}=\frac{\frac{1}{2}}{1-\frac{x}{2}}$ This is now a geometric series with first term $a=\dfrac{1}{2}\,$ and common ratio $r=\frac{x}{2}\,$. Therefore, the series is as follows. $ \frac{1}{2}+\frac{x}{4}+\frac{x^2}{8}+\ldots +\frac{x^n}{2^{n+1}}+\ldots $